Find $\lim_{x\to 0}\dfrac{e^x-\cos{x}}{4\sin{x}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac14$ (Choice B) B $\dfrac{e-1}{4}$ (Choice C) C $-\dfrac12$ (Choice D) D The limit doesn't exist.
Explanation: Substituting $x=0$ into $\dfrac{e^x-\cos{x}}{4\sin{x}}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 0}\dfrac{e^x-\cos{x}}{4\sin{x}} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[e^x-\cos{x}]}{\dfrac{d}{dx}[4\sin(x)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{e^x+\sin(x)}{4\cos(x)} \\\\ &=\dfrac{e^{(0)}+\sin(0)}{4\cos(0)} \gray{\text{Substitution}} \\\\ &=\dfrac14 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[e^x-\cos{x}]}{\dfrac{d}{dx}[4\sin(x)]}$ actually exists. In conclusion, $\lim_{x\to 0}\dfrac{e^x-\cos{x}}{4\sin{x}}=\dfrac14$.